Đáp án:
\(\begin{array}{l}
d,\,\,\,\, - 2\\
e,\,\,\,\,\,2\sqrt 3 - 6\\
f,\,\,\,\,3
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
d,\\
\dfrac{6}{{2 - \sqrt {10} }} + \dfrac{{5\sqrt 2 - 2\sqrt 5 }}{{\sqrt 5 - \sqrt 2 }}\\
= \dfrac{{6.\left( {2 + \sqrt {10} } \right)}}{{\left( {2 - \sqrt {10} } \right)\left( {2 + \sqrt {10} } \right)}} + \dfrac{{{{\sqrt 5 }^2}.\sqrt 2 - {{\sqrt 2 }^2}.\sqrt 5 }}{{\sqrt 5 - \sqrt 2 }}\\
= \dfrac{{6.\left( {2 + \sqrt {10} } \right)}}{{{2^2} - {{\sqrt {10} }^2}}} + \dfrac{{\sqrt 5 .\sqrt 2 .\left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt 5 - \sqrt 2 }}\\
= \dfrac{{6.\left( {2 + \sqrt {10} } \right)}}{{ - 6}} + \sqrt 5 .\sqrt 2 \\
= - \left( {2 + \sqrt {10} } \right) + \sqrt {10} \\
= - 2\\
e,\\
\dfrac{2}{{\sqrt 3 + 1}} - \dfrac{{5\sqrt 3 - 3}}{{\sqrt 3 - 5}} - 5\\
= \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} - \dfrac{{5\sqrt 3 - {{\sqrt 3 }^2}}}{{\sqrt 3 - 5}} - 5\\
= \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{{{{\sqrt 3 }^2} - {1^2}}} - \dfrac{{\sqrt 3 .\left( {5 - \sqrt 3 } \right)}}{{\sqrt 3 - 5}} - 5\\
= \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{2} - \left( { - \sqrt 3 } \right) - 5\\
= \left( {\sqrt 3 - 1} \right) + \sqrt 3 - 5\\
= 2\sqrt 3 - 6\\
f,\\
\dfrac{{2\sqrt 3 + 3}}{{\sqrt 3 }} - \dfrac{2}{{\sqrt 3 + 1}}\\
= \dfrac{{2\sqrt 3 + {{\sqrt 3 }^2}}}{{\sqrt 3 }} - \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\\
= \dfrac{{\sqrt 3 .\left( {2 + \sqrt 3 } \right)}}{{\sqrt 3 }} - \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{{{{\sqrt 3 }^2} - {1^2}}}\\
= \left( {2 + \sqrt 3 } \right) - \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{2}\\
= \left( {2 + \sqrt 3 } \right) - \left( {\sqrt 3 - 1} \right)\\
= 3
\end{array}\)
